indoarts
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December 16, 2021, 06:48:38 AM |
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noobs question and Just wondering , if you find the private key from one of those private keys puzzle , you just input the pv key to core wallet , or another wallet , or there is another step we should do l?
That's it, you just import the private key in WIF to core wallet and the BTC are yours. Cheers. thank you for your kind reply cheers
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Texgen
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December 18, 2021, 05:35:51 PM |
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Hello everyone, have you checked the open puzzles for the presence of coins in other networks? There is a token on the 115th puzzle on the BSC network, can it be taken away?
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PrivatePerson
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December 20, 2021, 09:54:39 AM |
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Hello everyone, have you checked the open puzzles for the presence of coins in other networks? There is a token on the 115th puzzle on the BSC network, can it be taken away?
xenon? its scamcoin.
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Alpaste
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December 27, 2021, 08:47:48 PM |
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Hello guys, i've been trying to randomly bruteforce puzzle #64 for WEEKS in a row. I was using Bitcoincrack to randomly Bruteforce addresses that share the same 7 letters of 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN. after bitcoincrack generated today 1000 addresses that share the same first 7 address characters, i have always looked for the last 3 characters of the generated addresses to go through the addresses quickly and see if #64 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN is on the list. and indeed, i saw an addresses that ends with the last 3 characters "XQN" that the original puzzle 64 address have. And the moment i saw XQN, i freaking screamed I WON 30K EUROS and smashed couple of glass bottles and went quickly to say and to celebrate with my friends. after my friends came, we checked the address again and it was NOT the address i was looking for. it was so embarrassment, and i'm now broken and sad as hell. What are the odds that i find the last 3 characters address, that i'm looking for in the range 64? ITS pretty much low, that's why i thought i found the 0.64 BTC. but unfortunately not. I should be the unluckiest guy ever... Here is the Address, i thought i got the 0.64 bitcoin from it. isn't that sad situation? Pub Addr: 16jY7qLHKTofWA2DYS9ZaWyvZ5qnjfSXQN Priv (WIF): p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZ3tw8CPZjZciVwEz8EV Priv (HEX): 0x000000000000000000000000000000000000000000000000DAA7B7BDF277A7B7
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LoyceV
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Thick-Skinned Gang Leader and Golden Feather 2021
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December 28, 2021, 01:50:38 PM |
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I was using Bitcoincrack to randomly Bruteforce addresses that share the same 7 letters of 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN. after bitcoincrack generated today 1000 addresses that share the same first 7 address characters ~ and indeed, i saw an addresses that ends with the last 3 characters "XQN" that the original puzzle 64 address have. ~ What are the odds that i find the last 3 characters address Each character has 58 possibilities, so if you have 1000 addresses, the chance of finding the same 3 characters at the end is about 0.5%. However, you need to find 24 more characters, and those aren't magically going to match once you find the last 3.
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itod
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^ Will code for Bitcoins
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December 28, 2021, 02:08:34 PM |
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Hello guys, i've been trying to randomly bruteforce puzzle #64 for WEEKS in a row. ... after my friends came, we checked the address again and it was NOT the address i was looking for. it was so embarrassment, and i'm now broken and sad as hell. What are the odds that i find the last 3 characters address, that i'm looking for in the range 64? ITS pretty much low, that's why i thought i found the 0.64 BTC. but unfortunately not. I should be the unluckiest guy ever...
Everyone is brute forcing #64 since more than a two years now. You were very, very, optimistic that you will be the one with the stroke of unbelivable luck. Look at it this way: You uwere really lucky to understand after only few weeks how small chances are that this will happen and you can now move to more interesting things.
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willi9974
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December 29, 2021, 02:26:23 AM Last edit: December 29, 2021, 02:20:02 PM by willi9974 |
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have at the moment time to try my luck. what is the best way to start the searching for puzzle 64 and what for a script or program should i use for that? a pc with 3 3070er GPU are here... can anyone help to start? Best regards Willi Update: For the german Community i openenddiskussion a new german topic —> https://bitcointalk.org/index.php?topic=5379045.0
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PrivatePerson
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December 29, 2021, 03:06:10 PM |
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willi9974 Have you read what is written above your post? You are apparently not too smart for this puzzle because in your topic you include a link to an old topic when there is a new, relevant one. First, learn to read, then think about the puzzle.
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willi9974
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Escrow Service
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December 29, 2021, 03:11:25 PM |
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Thanks i will follow the new one ;-)
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MrFreeDragon
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January 04, 2022, 12:10:42 PM |
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I was using Bitcoincrack to randomly Bruteforce addresses that share the same 7 letters of 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN. after bitcoincrack generated today 1000 addresses that share the same first 7 address characters ~ and indeed, i saw an addresses that ends with the last 3 characters "XQN" that the original puzzle 64 address have. ~ What are the odds that i find the last 3 characters address Each character has 58 possibilities, so if you have 1000 addresses, the chance of finding the same 3 characters at the end is about 0.5%. However, you need to find 24 more characters, and those aren't magically going to match once you find the last 3. He actually found 9 characters (10 characters including the 1st "1"): 16jY7qLJnxb7CHZyqBP8qca9d51gAjy XQN - #64 puzzle 16jY7qLHKTofWA2DYS9ZaWyvZ5qnjfS XQN - found by Alpaste But looks nice to have a vanity address with 10 characters and the same bit size.
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fxsniper
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January 05, 2022, 01:27:30 AM |
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bitCrack scan every single address pro - scan every address cons - use too much time for scan reference from result https://hashkeys.club/64/results/Just an idea may be need to modify new one bitCrack engine to JumpCrack or SkipCrack or PatternCrack like to scan and jump to other number but it is can possible to missing that address found 100% but some one choose right jump maybe lucky range 64 bit is 18446744073709551616 if jump every trillion it can be scan all easy (yes with missing address) python code will be easy to code for scan by jump but will be very slow CUDA GPU scan will be answer for jump scan may be jump for 43 bit range and scan all on 32 bit range I think this idea may be some one can found key #64 just idea and it possible missing address 100%
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Kostelooscoin
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January 07, 2022, 03:55:05 PM |
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bitCrack scan every single address pro - scan every address cons - use too much time for scan reference from result https://hashkeys.club/64/results/Just an idea may be need to modify new one bitCrack engine to JumpCrack or SkipCrack or PatternCrack like to scan and jump to other number but it is can possible to missing that address found 100% but some one choose right jump maybe lucky range 64 bit is 18446744073709551616 if jump every trillion it can be scan all easy (yes with missing address) python code will be easy to code for scan by jump but will be very slow CUDA GPU scan will be answer for jump scan may be jump for 43 bit range and scan all on 32 bit range I think this idea may be some one can found key #64 just idea and it possible missing address 100% hello can you share your modified version of bitcrack please ?
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dextronomous
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January 09, 2022, 01:14:08 AM |
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something this damn shit doesn't want to give up ![Grin](https://bitcointalk.org/Smileys/default/grin.gif) maybe we can try to get in through ripemd160... These stories abour lot's of BTC being lost are just legends "from Russia with love" Hi there Andzhig, you have a complete package of these things you are programming to start searching somehow. thanks man,
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Feron
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January 12, 2022, 11:46:47 AM Last edit: January 12, 2022, 12:09:17 PM by Feron |
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Random base58 happy hunting code: import base58 import secrets from bit import * for x in range(100000000): c0 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c1 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c2 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c3 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c4 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c5 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c6 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c7 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c8 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c9 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") cc = ("NPQRSTUVWXYZabcdefghij") for x0 in cc: k1 = Key.from_hex(base58.b58decode(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9).hex()) ad = k1.address if ad == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN": f=open("xxx.txt","a") f.write(str(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9)+"-"+(ad)+"\n") f.close() print(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9,ad,x)
New version improve speed code: import base58 import secrets from bit import * for x in range(100000000): c0 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c1 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c2 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c3 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c4 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c5 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c6 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c7 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c8 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") c9 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz") cc = ("NPQRSTUVWXYZabcdefghij") for x0 in cc: k1 = Key.from_hex(base58.b58decode(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9).hex()) if (str(k1)).endswith("XQN>"): print(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9,k1,x) if (str(k1)) == "<PrivateKey: 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN>": f=open("win.txt","a") f.write(str(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9)+(str(k1))+"\n") f.close()
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Feron
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January 13, 2022, 06:05:38 PM Last edit: January 13, 2022, 09:47:13 PM by Mr. Big |
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New test code you can customize it to suit your needs puzle address 27 test code : from bit import Key for xx in range(1,100000): for x in range(1,100000): y = 1 p = 1 p = (x - y) / xx # 100000 # 65536 y += 1 p *= p pp = 1 - p if pp >= 0.0: # 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 c1 = ''.join(str(pp))[11:] for c in ("11",): ke = Key.from_int(int(c+c1)) if (str(ke)).endswith("86k>"): print(c+c1,ke,x,xx) if (str(ke)) == "<PrivateKey: 128z5d7nN7PkCuX5qoA4Ys6pmxUYnEy86k>": f=open("win.txt","a") f.write(str(c+c1)+(str(ke))+"\n") f.close() interesting for this collision is this number 1892x,,,3187xx I don't know what that means but it doesn't matter for fun
Hi all this easy matematic 1+1 code code: from bit import Key import random for x in range(1000000000): xx = random.randint(4611686018427387904,9223372036854775808) cc = xx+xx k1 = Key.from_int(int(cc)) if (str(k1)).startswith("<PrivateKey: 16jY"): print(cc,xx,k1,x) if (str(k1)) == "<PrivateKey: 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN>": f=open("win.txt","a") f.write(str(cc)+(str(k1))+"\n") f.close() the simpler the faster code compute 4611686018427387904 to 9223372036854775808 take one number maximum 9223372036854775808 keyspace and doubles it him final decimal is private key little reduces the key space
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Feron
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January 14, 2022, 07:50:52 PM |
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it seems to me that 64 is not in this range because if it had been there a long time ago, it would have been broken, so you better focus on 66 ddd this is just a theory that may be wrong Bitcrack use you use 8 rtx 3090, each gets 1 initial range 89abcdef and now you generate 3 + 3 hex + 1 = 7 hex one graphics card scans 9 hexadecimal 68719476735 which is 1 minute for rtx 3090 that's 1400 per day I tried it in python until the collision occurred at the beginning sometimes once every 1,000,000, etc.
This method is risky, it is better to mine ethereum or other profitable coins
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Feron
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January 14, 2022, 08:39:16 PM |
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maybe it sounds funny ddd but this interests me does not block writing and printing of the bit library if there is a collision of addresses with the balance or there is no protection against brute force ![Grin](https://bitcointalk.org/Smileys/default/grin.gif) dddd ![Grin](https://bitcointalk.org/Smileys/default/grin.gif) some programmer could check it and let me know
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Feron
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January 15, 2022, 01:08:39 AM Last edit: January 15, 2022, 09:14:49 AM by Feron |
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shishua random generator for fun run faster use cython any codes important all addresses are generated via x sequence you can't start from the beginning because they would be repeated it's too small save have fun example I ended up at 372055 according to print you will know where you ended up and just rewrite it to your number and save the progress you can use the words of the alphabet ascii etc,i used numbers from shishua import SHISHUA from bit import Key for x in range(372055,10000000): rng = SHISHUA(x).random_raw(7).hex() zzz = ["80","81","82","83","84","85","86","87","88","89","8a","8b","8c","8d","8e","8f","90","91","92","93","94","95","96","97","98","99","9a","9b","9c","9d","9e","9f","a0","a1","a2","a3","a4","a5","a6","a7","a8","a9","aa","ab","ac","ad","ae","af","b0","b1","b2","b3","b4","b5","b6","b7","b8","b9","ba","bb","bc","bd","be","bf","c0","c1","c2","c3","c4","c5","c6","c7","c8","c9","ca","cb","cc","cd","ce","cf","d0","d1","d2","d3","d4","d5","d6","d7","d8","d9","da","db","dc","dd","de","df","e0","e1","e2","e3","e4","e5","e6","e7","e8","e9","ea","eb","ec","ed","ee","ef","f0","f1","f2","f3","f4","f5","f6","f7","f8","f9","fa","fb","fc","fd","fe","ff"] for c in zzz: ke1 = Key.from_hex(c+rng) ad1 = ke1.address if ad1.endswith("XQN"): print(c+rng,ad1,x) if ad1 == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN": f=open("win.txt","a") f.write(str(c+rng)+"-"+(ad1)+"\n") f.close()
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DeepComplex
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January 15, 2022, 05:58:02 AM |
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shishua random generator for fun run faster use cython any codes from shishua import SHISHUA from bit import Key for x in range(372055,10000000): rng = SHISHUA(x).random_raw(7).hex() zzz = ["80","81","82","83","84","85","86","87","88","89","8a","8b","8c","8d","8e","8f","90","91","92","93","94","95","96","97","98","99","9a","9b","9c","9d","9e","9f","a0","a1","a2","a3","a4","a5","a6","a7","a8","a9","aa","ab","ac","ad","ae","af","b0","b1","b2","b3","b4","b5","b6","b7","b8","b9","ba","bb","bc","bd","be","bf","c0","c1","c2","c3","c4","c5","c6","c7","c8","c9","ca","cb","cc","cd","ce","cf","d0","d1","d2","d3","d4","d5","d6","d7","d8","d9","da","db","dc","dd","de","df","e0","e1","e2","e3","e4","e5","e6","e7","e8","e9","ea","eb","ec","ed","ee","ef","f0","f1","f2","f3","f4","f5","f6","f7","f8","f9","fa","fb","fc","fd","fe","ff"] for c in zzz: ke1 = Key.from_hex(c+rng) ad1 = ke1.address if ad1.endswith("XQN"): print(c+rng,ad1,x) if ad1 == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN": f=open("win.txt","a") f.write(str(c+rng)+"-"+(ad1)+"\n") f.close() @Feron How can I run this in Ubuntu? Can I use pip to install the shishua python package?
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